Uranium-238 (atomic number 92) decays in a series to lead-206 (atomic number 82)…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
Uranium-238 (atomic number ) decays in a series to lead-206 (atomic number ). Thorium-232 (atomic number ) decays to lead-208. In each series only alpha particles and beta-minus particles are emitted. Find the number of alpha particles and the number of beta-minus particles in each series.
Need a reminder? The key ideaShow
In every decay the total mass number and the total charge stay the same. An alpha decay changes by . A beta-minus decay changes it by . The energy released is , with atomic masses: for alpha decay, for beta-minus, and for beta-plus, where MeV. The products share the kinetic energy in the inverse ratio of their masses.
The worked solution
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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Plan. The mass number fixes the number of alpha particles. The atomic number then fixes the number of beta particles.
Step 1. Uranium-238. The mass number falls by , so the number of alpha particles is . They lower the atomic number by , from to . Lead has , so beta-minus particles have each put back one proton.
Step 2. Thorium-232. The mass number falls by , so there are alpha particles. The atomic number goes from to . To reach we need beta-minus particles.
Answer: Uranium-238: alpha and beta. Thorium-232: alpha and beta.
Check. For uranium, . For thorium, . Both end at lead.
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